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Using Green's theorem to evaluate

  • Thread starter DryRun
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  • #1
DryRun
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Homework Statement
Using Green's theorem, evaluate:
http://s2.ipicture.ru/uploads/20120117/6p57O2HO.jpg

The attempt at a solution
[tex]\frac{\partial P}{\partial y}=3x+2y[/tex]
[tex]\frac{\partial Q}{\partial x}=2y+10x[/tex]
[tex]\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=7x[/tex]
To do the integration, i'm using the parametric equations of the circle; x= cosθ and y=sinθ
[tex]\int\int 7\cos \theta \,.rdrd\theta[/tex]
The curve C is a circle with centre (1,-2) and radius r=1
I've drawn the graph in my copybook.

Description of circle:
For θ fixed, r varies from r=-2 to r=-3
θ varies from -∏/2 to -∏/2

However, i can't get the correct answer as i'm quite sure that i've got the wrong limits. Usually, i deal with circles with centre (0,0) so it's easier to find the limits for r and θ.
 

Answers and Replies

  • #2
DryRun
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Any help?
 
  • #3
LCKurtz
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Homework Statement
Using Green's theorem, evaluate:
http://s2.ipicture.ru/uploads/20120117/6p57O2HO.jpg

The attempt at a solution
[tex]\frac{\partial P}{\partial y}=3x+2y[/tex]
[tex]\frac{\partial Q}{\partial x}=2y+10x[/tex]
[tex]\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=7x[/tex]
To do the integration, i'm using the parametric equations of the circle; x= cosθ and y=sinθ
[tex]\int\int 7\cos \theta \,.rdrd\theta[/tex]
The curve C is a circle with centre (1,-2) and radius r=1
I've drawn the graph in my copybook.

Description of circle:
For θ fixed, r varies from r=-2 to r=-3
θ varies from -∏/2 to -∏/2

However, i can't get the correct answer as i'm quite sure that i've got the wrong limits. Usually, i deal with circles with centre (0,0) so it's easier to find the limits for r and θ.
Your equations ##x=r\cos\theta,\ y=r\sin\theta## give a circle at the origin, not centered at ##(1,-2)##. First figure out how to fix that (think about translation). Then, standard convention in polar coordinates uses positive r, so it wouldn't go from -2 to -3. And of course if you use the same upper and lower limits for ##\theta##, you will get 0. Think about if ##
\theta## starts at ##-\frac \pi 2## and increases by ##2\pi##, where will it end?
 
  • #4
DryRun
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Hi LCKurtz

Here is the Cartesian graph of path C:

http://s1.ipicture.ru/uploads/20120118/SQzbOw3U.gif

An attempt at translation of axes:

Let x = X+h
Let y=Y+k
Then, X=rcosθ-h and Y=rsinθ-k

Here, x=1 and y=-2.
I'm not sure. Is this correct?

I'm going to try another method:
From the general equation for polar coordinates, [itex]x^2+y^2=r^2[/itex], at (1,-2), r=√5
This would mean that r varies from r=0 to r=√5
Then, at (1,-2), x=1 and y=-2
Substituting into x=rcosθ and y=rsinθ
1=√5cosθ, θ= 1.107
and -2=√5sinθ, θ=-1.107
Thus θ varies from θ=-1.107 to θ=1.107
Again, i'm not sure. Maybe there's a simpler and more accurate method?

I did some more research and i believe this method is correct:
The new centre of circle is: [(x-1),(y+2)] and the radius stays same, r=1.
For [itex]x=r\cos\theta,\ y=r\sin\theta[/itex], these become [itex](x-1)=r\cos\theta,\ (y+2)=r\sin\theta[/itex]
So, [itex]x=r\cos\theta+1,\ y=r\sin\theta-2[/itex]
Since, r=1, the parametric equations of the circle become: [itex]x=\cos\theta+1,\ y=\sin\theta-2[/itex]

Therefore, the integral becomes:
[tex]\int\int 7(\cos\theta+1) \,.rdrd\theta=\int\int 7\cos\theta+7 \,.rdrd\theta[/tex]
The description of the circle:
For θ fixed, r varies from r=0 to r=1.
θ varies from θ=0 to θ=2∏.

The integration gives: 7∏, which is the correct answer. So, i assume that i finally found the right method.
 
Last edited:
  • #5
LCKurtz
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So, [itex]x=r\cos\theta+1,\ y=r\sin\theta-2[/itex]
Since, r=1, the parametric equations of the circle become: [itex]x=\cos\theta+1,\ y=\sin\theta-2[/itex]

Therefore, the integral becomes:
[tex]\int\int 7(\cos\theta+1) \,.rdrd\theta=\int\int 7\cos\theta+7 \,.rdrd\theta[/tex]
The description of the circle:
For θ fixed, r varies from r=0 to r=1.
θ varies from θ=0 to θ=2∏.

The integration gives: 7∏, which is the correct answer. So, i assume that i finally found the right method.
Yes! Good work, and you learned something in the process. :biggrin:
 
  • #6
DryRun
Gold Member
838
4
Thanks for your guidance, LCKurtz. It's much appreciated.:smile:
 

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